A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^{1} 2 V / m s .$ what is the displacement current?
1.39 $A$
13.9 $A$
0.07 $A$
139 $A$

Question

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^{1} 2 V / m s .$ what is the displacement current?
1.39 $A$
13.9 $A$
0.07 $A$
139 $A$

Toppr Admin · Accepted Answer

Id-x2208-0d-x3D5-dt-x2208-0d-EA-dt-x2208-0AEAdt-x2208-x3C0-r2dEdt-8-85-xD7-10-x2212-12-xD7-3-14-xD7-25-xD7-10-x2212-4-xD7-1012-0-07AHence-option -A- is correct answer-

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of 1012V/ms. what is the displacement current?1.39A13.9A0.07A139A

Id=∈0dϕdt=∈0d(EA)dt=∈0AEAdt=∈πr2dEdt=8.85×10−12×3.14×25×10−4×1012=0.07AHence,option (A) is correct answer.

A parallel plate condenser has two circular metal plates of radius 15 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of $10^{1} 2 V / m s .$ what is the displacement current?
1.39 $A$
13.9 $A$
0.07 $A$
139 $A$

$\begin{matrix} I d = \in_{0} \frac{d ϕ}{d t} = \in_{0} \frac{d (E A)}{d t} = \in_{0} A \frac{E A}{d t} =\in \frac{π r^{2} d E}{d t} = 8.85 \times 10^{- 12} \times 3.14 \times 25 \times 10^{- 4} \times 10^{12} = 0.07 A \end{matrix}$
Hence,
option $(A)$ is correct answer.